Question:

Let \( \{N(t)\}_{t \geq 0} \) be a Poisson process with rate 1. Consider the following statements.
[(I)] \( P(N(3) = 3 \mid N(5) = 5) = \binom{5}{3} \left(\frac{3}{5}\right)^3 \left(\frac{2}{5}\right)^2 \).
[(II)] If \( S_5 \) denotes the time of occurrence of the 5th event for the above Poisson process, then \( E(S_5 \mid N(5) = 3) = 7 \).
Which of the above statements is/are true?

Show Hint

- For a Poisson process, conditional distributions of event counts in disjoint intervals are binomial.
- The expected time of the $n$th event in a Poisson process is the sum of $n$ independent exponential random variables, resulting in a Gamma distribution.
Updated On: Aug 30, 2025
  • Only (I)
  • Only (II)
  • Both (I) and (II)
  • Neither (I) nor (II)
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The Correct Option is C

Solution and Explanation

1) Analyzing statement (I):
For a Poisson process, given the event $N(5) = 5$, the distribution of $N(3)$ is binomial, as the number of events occurring in the first 3 units of time, given the total number of events in 5 units, follows a binomial distribution. The probability of $N(3) = 3$ given $N(5) = 5$ is: \[ P(N(3) = 3 \mid N(5) = 5) = \binom{5}{3} \left(\frac{3}{5}\right)^3 \left(\frac{2}{5}\right)^2. \] Thus, statement (I) is correct. 
2) Analyzing statement (II):
The time of occurrence of the 5th event, $S_5$, in a Poisson process with rate 1 follows a Gamma distribution with shape parameter 5 and rate 1. The expected value of $S_5$, given that there are 3 events by time 5, is: \[ E(S_5 \mid N(5) = 3) = 7. \] Thus, statement (II) is also correct. Therefore, both statements (I) and (II) are true, and the correct answer is (C).

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